The expressions which represents the slope of a tangent to the curve [tex]y=\frac{1}{x+1}[/tex] at any point (x, y) are:
[tex]f'(x) =limh \rightarrow 0\frac{-h}{h(x+1)(x+h+1)}\\\\f'(x) =limh \rightarrow 0\frac{-h}{x^2h+2xh+xh^2+h^2 +1}[/tex]
The slope of a tangent to the curve.
Mathematically, the slope of a tangent line to the curve is given by this equation:
[tex]f'(x) =limh \rightarrow 0\frac{f(x+h)-f(x)}{h}[/tex]
Given the function:
[tex]f(x)=y=\frac{1}{x+1}[/tex]
When (x + h), we have:
[tex]f(x+h)=y=\frac{1}{x+h+1}[/tex]
Next, we would find the derivative of f(x):
[tex]f'(x) =limh \rightarrow 0\frac{\frac{1}{x+h+1} -\frac{1}{x+1}}{h}\\\\f'(x) =limh \rightarrow 0\frac{x+1 -(x+h+1)}{h(x+1)(x+h+1)}\\\\f'(x) =limh \rightarrow 0\frac{x+1 -x-h-1}{h(x+1)(x+h+1)}\\\\f'(x) =limh \rightarrow 0\frac{-h}{h(x+1)(x+h+1)}\\\\f'(x) =limh \rightarrow 0\frac{-h}{x^2h+2xh+xh^2+h^2 +1}\\\\f'(x) = \frac{-1}{(x+1)(x+1)} \\\\f'(x) = \frac{-1}{(x+1)^2}[/tex]
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